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blagie [28]
3 years ago
5

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.

8�107m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 2.2�1011m from its star with a period of 402 earth days. Once on the surface you find that the free-fall acceleration is 12.2m/s2.
a. what is the mass of the Planet?
b. what is the mass of the star?
Physics
1 answer:
trapecia [35]3 years ago
3 0

Answer

given,

diameter of planet = 1.8 x 10⁷ m

radius of planet = 0.9 x 10⁷ m

time period = 22.3 hours

the planet orbits 2.2 x 10¹¹ m   period of 402 earth days.

acceleration= 12.2 m/s²

we know

g = \dfrac{GM_{p}}{r^2}

M_{p} = \dfrac{gr^2}{G}

M_{p} = \dfrac{12.2 \times (0.9 \times 10^7)^2}{6.67 \times 10^{-11}}

M_p = 1.48 x 10²⁵  Kg

b) Formula to calculate the mass of star

M_{s} = \dfrac{4\pi^2}{GT_s^2}(R^3)

M_{s} = \dfrac{4\pi^2}{6.67 \times 10^{-11} \times (402 \times 86400)^2}((2.2 \times 10^{11})^3)

M_s = 5.22 x 10³³ Kg

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The slope represents the acceleration
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An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t
viktelen [127]

Answer:

a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

          F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0
3 years ago
A pitcher exerts a force on a baseball that is 30 times the balls weight. How fast is the pitcher accelerating the ball?
iVinArrow [24]

Answer:

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  We have force applied is also equal to product of mass and acceleration.

                            F = ma = 30 x mg

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Answer:

c

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So the option c is correct.

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